Respuesta :
Answer:
[tex] z = \frac{2-3}{\frac{1}{\sqrt{10}}}=-3.163[/tex]
[tex] z = \frac{4-3}{\frac{1}{\sqrt{10}}}=3.163[/tex]
[tex]P(-3.163<z<3.163)=P(z<3.163)-P(z<3.163)[/tex]
[tex]P(-3.163<z<3.163)=P(z<3.163)-P(z<3.163)= 0.9992 -0.00078= 0.99842[/tex]
So then we will expect 9.98 packages between 2-4 rookie cards in the sample of 10
Step-by-step explanation:
Let X the random variable that represent the number of rookie cards of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(3,1)[/tex]
Where [tex]\mu=3[/tex] and [tex]\sigma=1[/tex]
We select a sample size of n = 10 variety packs and we want to find this probability:
[tex]P(2<\bar X<4)[/tex]
We can use the z score formula given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
We can find the z score for 2 and 4 and we got:
[tex] z = \frac{2-3}{\frac{1}{\sqrt{10}}}=-3.163[/tex]
[tex] z = \frac{4-3}{\frac{1}{\sqrt{10}}}=3.163[/tex]
So we can find the probability with this difference
[tex]P(-3.163<z<3.163)=P(z<3.163)-P(z<3.163)[/tex]
And using the normal standard distirbution or excel we got:
[tex]P(-3.163<z<3.163)=P(z<3.163)-P(z<3.163)= 0.9992 -0.00078= 0.99842[/tex]
So then we will expect 9.98 packages between 2-4 rookie cards in the sample of 10
